The integrating factor of the differential eq | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given the differential equation: $\frac{dy}{dx} = -(3x^2 	an^{-1} y - x^3)(1 + y^2)$.Rearranging the terms: $\frac{dy}{dx} = x^3(1 + y^2) - 3x^2(

Vedclass · Accepted Answer

$e^{x^3}$

Vedclass · Answer

(B) Given the differential equation: $\frac{dy}{dx} = -(3x^2 	an^{-1} y - x^3)(1 + y^2)$.Rearranging the terms: $\frac{dy}{dx} = x^3(1 + y^2) - 3x^2(	an^{-1} y)(1 + y^2)$.Dividing both sides by $(1 + y^2)$: $\frac{1}{1 + y^2} \cdot \frac{dy}{dx} = x^3 - 3x^2 	an^{-1} y$.Rearranging to standard form: $\frac{1}{1 + y^2} \cdot \frac{dy}{dx} + 3x^2 	an^{-1} y = x^3$.Let $t = 	an^{-1} y$, then $\frac{dt}{dx} = \frac{1}{1 + y^2} \cdot \frac{dy}{dx}$.The equation becomes: $\frac{dt}{dx} + 3x^2 t = x^3$.This is a linear differential equation of the form $\frac{dt}{dx} + P(x)t = Q(x)$, where $P(x) = 3x^2$.The integrating factor $(IF)$ is given by $e^{\int P(x) dx} = e^{\int 3x^2 dx} = e^{x^3}$.

The integrating factor of the differential equation $\frac{dy}{dx} + (3x^2 \tan^{-1} y - x^3)(1 + y^2) = 0$ is

Similar Questions

Let $y=y(x)$ be the solution of the differential equation $x\frac{dy}{dx}-y=x^{2}\cot x, x\in(0,\pi)$. If $y(\frac{\pi}{2})=\frac{\pi}{2}$, then $6y(\frac{\pi}{6})-8y(\frac{\pi}{4})$ is equal to :

If $x^2 y - x^3 \frac{dy}{dx} = y^4 \cos x$,then $x^3 y^{-3}$ is equal to

The integrating factor of the differential equation $x \frac{dy}{dx} - y = x^3, (x > 0)$ is . . . . . . .

The general solution of $\frac{dy}{dx} + y \tan x = 2x + x^2 \tan x$ is:

If $y(x)$ satisfies the differential equation $y^{\prime}-y \tan x=2 x \sec x$ and $y(0)=0$,then which of the following is true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module